Consider the given function \(𝑓(𝑥)\). \(f(x) =\begin{cases} ax + b, &…
2025
Consider the given function \(𝑓(𝑥)\).
\(f(x) =\begin{cases} ax + b, & \text{for } x < 1 \\x^3 + x^2 + 1, & \text{for } x \geq 1\end{cases}\)
If the function is differentiable everywhere, the value of 𝑏 must be ________. (rounded off to one decimal place)
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Correct answer: -2
For f to be differentiable everywhere, it must be continuous at x = 1 and the left and right derivatives at x = 1 must be equal.
Continuity at x = 1: the left-hand value is a(1) + b = a + b, and the right-hand value is 1^3 + 1^2 + 1 = 3. Therefore a + b = 3, so b = 3 - a.
Differentiability at x = 1: the derivative from the left is a. The derivative from the right of x^3 + x^2 + 1 is 3x^2 + 2x, which at x = 1 equals 3 + 2 = 5. Thus a = 5.
Substitute a = 5 into b = 3 - a to get b = 3 - 5 = -2, which rounded to one decimal place is -2.0.
Answer: b = -2.0
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